Three Dimensional Geometry

$ 0=\frac{6+3 \alpha}{2+3} \Rightarrow \alpha=-2 $

$ \begin{aligned} &0=\frac{4 \beta+20}{9} \Rightarrow \beta=-5\\ &\text { }\\ &\begin{aligned} So,\,\,& A \equiv(-2,4,7) \\ & \beta \equiv(3,-5,8) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{A C}{B C}=\frac{2}{5} \\ & C_x=\frac{2 \times 3-5 \times(-2)}{2-5}=\frac{-16}{3} \\ & C_y=\frac{-10-20}{2-5}=10 \\ & C_z=\frac{2 \times 8-5 \times 7}{-3}=\frac{19}{3} \end{aligned}\\ &\text { So, } \quad C \equiv\left(-\frac{16}{3}, 10, \frac{19}{3}\right) \end{aligned} $

Let direction cosine of $X$-axis be ( $1,0,0$ )

direction cosine of line having direction ratio as $3,-1,5$

$ 1= \pm \frac{3}{\sqrt{35}}, m=\mp-\frac{1}{\sqrt{35}}, n= \pm \frac{5}{\sqrt{35}} $

Direction cosine of line bisecting the angle below two given lines is

$ \begin{aligned} \left(l_1 \pm l_2, m_1 \pm m_2, n_1 \pm n_2\right) & \\ \left(\frac{\sqrt{35}-3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}}\right) \\ \left(\frac{\sqrt{35}-3}{\sqrt{5}}, \frac{1}{\sqrt{5}},-\sqrt{5}\right) \end{aligned} $

Equation of plane $\pi_1$ :

$ 2 x+y-z-\frac{\alpha}{2}=0 $

Equation of plane $\pi_2: 2 x+y-z+1=0$

Distance below plane $\pi_1$ and plane $\pi_2$

$ \begin{aligned} & =\frac{\left|1+\frac{\alpha}{2}\right|}{\sqrt{2^2+1^2+1^2}}=2=\left|1+\frac{\alpha}{2}\right|=2 \sqrt{6} \\ & \Rightarrow \frac{\alpha}{2}+1= \pm 2 \sqrt{6} \\ & \Rightarrow \alpha+2= \pm 4 \sqrt{6} \\ & \Rightarrow \alpha=-2+4 \sqrt{6} \text { and } \alpha=-2-4 \sqrt{6} \\ & \text { Product }=(-2+4 \sqrt{6})(-2-4 \sqrt{6}) \\ & =4-96=-92 \end{aligned} $

Let coordinates of point be $(x, y, z)$.

Distance from $X Y$ plane $=2 \times$ Distance from $Z$ axis.

$ \begin{aligned} & |z|=2 \times \sqrt{x^2+y^2} \\ \Rightarrow & z^2=4 x^2+4 y^2 \Rightarrow 4 x^2+4 y^2-z^2=0 \end{aligned} $

Let side of cube be $a$.

Let coordinates of vertices of cube are shown as in figure.

Ratio direction ratio of line $A Q$ is $a: a:-a$

Ratio of direction ratio of line $B R$ is $a:-a: a$.

$ \cos \alpha=\frac{\left|a^2-a^2-a^2\right|}{\sqrt{3} a \times \sqrt{3} a}=\frac{1}{3} $

Let ratio of direction ratio of line $B D$ be $a:-a: 0$

$ \begin{array}{ll} & \cos \beta=\frac{a^2+a^2}{\sqrt{3} a \times \sqrt{2} a}=\sqrt{\frac{2}{3}} \\ \Rightarrow & \cos \alpha+\cos ^2 \beta=\frac{1}{3}+\frac{2}{3}=1 \end{array} $

Let direction ratio of normal line of plane $\pi_1$ be $(a,-1,3)$.

Let direction ratio of normal line of plane $\pi_2$ be $(3, a, 1)$

$ \begin{aligned} & \cos 60^{\circ}=\frac{3 a-a+3}{\sqrt{a^2+10} \times \sqrt{a^2+10}}=\frac{2 a+3}{a^2+10} \\ & \Rightarrow \quad \frac{1}{2}=\frac{2 a+3}{a^2+10} \Rightarrow a^2+10=4 a+6 \\ & \Rightarrow \quad a^2-4 a+4=0 \Rightarrow a=2 \end{aligned} $

⇒ Direction ratio of line perpendicular to the plane $(a+2) x+(a-4) y+2 a z=a$

$ 4:-2: 4 $

Direction ratio 2: $-1: 2$

We have,

$ \begin{gathered} A(-4,9, k), B(-1,6, k) \text { and } C(0,7,10) \\ \mathbf{A B}=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+0 \hat{\mathbf{k}},|\mathbf{A B}|=3 \sqrt{2} \\ \mathbf{B C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+(10-k) \hat{\mathbf{k}} \\ |\mathbf{B C}|=\sqrt{1+1+(10-k)^2} \end{gathered} $

And $\mathbf{A C}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+(10-k) \hat{\mathbf{k}}$,

$ \begin{aligned} & |\mathbf{A C}|=\sqrt{16+4+(10-k)^2} \\ \therefore & |\mathbf{A C}|>|\mathbf{B C}| \\ \therefore & A C \rightarrow \text { Hypotenuse } \\ \therefore & \mathbf{A B} \cdot \mathbf{B C}=0 \\ \therefore & |\mathbf{A B}|=|\mathbf{B C}| \\ \Rightarrow & 18=2+(10-k)^2 \\ & 10-k= \pm 4 \Rightarrow k=10 \pm 4=14 \text { or } 6 \end{aligned} $

∴ No. of values of $k=2$

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{ll} & \alpha=60^{\circ}, \beta=45^{\circ}, \gamma=\theta \\ \because & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ \Rightarrow & \cos ^2 60+\cos ^2 45^{\circ}+\cos ^2 \theta=1 \\ \Rightarrow & \cos ^2 \theta=1-\frac{1}{4}-\frac{1}{2}=\frac{1}{4} \\ \Rightarrow & \cos \theta=\frac{1}{2} \Rightarrow \tan \theta=\sqrt{3} \end{array} \end{aligned} $

$ \begin{aligned} \mathbf{P Q}= & -\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \pi: a x+b y-3 z+d=0 \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \therefore & \mathbf{n}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ \therefore & \mathbf{P Q} \| \mathbf{n} \\ \therefore & \frac{-1}{a}=\frac{-2}{b}=\frac{3}{-3} \Rightarrow a=1, b=2 \end{array}\\ &\therefore Q(1,-2,0) \text { lies on plane }\\ &\begin{gathered} x+2 y-3 z+d, 1-4+d=0 \\ d=3 \\ a+b+d=1+2+3=6 \end{gathered} \end{aligned} $

For $\pi_1, n_1=a \times b$

where, $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$ \begin{aligned} \mathbf{b} & =\hat{\mathbf{i}}+\hat{\mathbf{k}} \\ \therefore \mathbf{n}_1 & =\left|\begin{array}{lll} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

For $\pi_2, \mathbf{n}_2=\mathbf{c} \times \mathbf{d}$ where $\mathbf{c}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$

$ \begin{aligned} \mathbf{d} & =\hat{\mathbf{k}}-\hat{\mathbf{i}} \\ \mathbf{n}_2 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Parallel vector line of intersection of $\pi_1$ and $\pi_2$

$ \begin{aligned} & \mathbf{a}=\mathbf{n}_1 \times \mathbf{n}_2 \\ & \mathbf{n}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{array}\right| \\ & \quad=\hat{\mathbf{i}}(-1+1)-\hat{\mathbf{j}}(1+1)+\hat{\mathbf{k}}(1+1) \\ & \quad=-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \begin{aligned} \text { And } \mathbf{b} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \therefore \cos \theta & =\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \\ \quad & =\left|\frac{-2-2}{\sqrt{4+4} \sqrt{3}}\right|=\left|\frac{4}{2 \sqrt{2} \sqrt{3}}\right| \\ \theta & =\cos ^{-1} \sqrt{\frac{2}{3}} . \end{aligned} \end{aligned} $

$ \text { Let } \frac{m}{n}=\frac{\lambda}{1} $

$ \begin{aligned} & \frac{p \lambda+2}{\lambda+1}=\frac{8}{5} \\ & 5 p \lambda+10=8 \lambda+8 \end{aligned} $

$ \begin{aligned} &\begin{gathered} 5 p \lambda-8 \lambda=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \lambda(8-5 p)=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \frac{-2 \lambda+p}{\lambda+1}=\frac{-1}{5} \\ -10 \lambda+5 p=-\lambda-1 \Rightarrow 5 p+1=9 \lambda \\ \lambda=\frac{5 p+1}{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ \frac{\lambda p+2}{\lambda+1}=\frac{8}{5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered}\\ &\text { Solving Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} & \left(\frac{5 p+1}{9}\right)(8-5 p)=2 \\ \Rightarrow & (5 p+1)(8-5 p)=18 \\ \Rightarrow & 40 p-25 p^2+8-5 p-18=0 \\ \Rightarrow & -25 p^2+35 p-10=0 \\ \Rightarrow & 5 p^2-7 p+2=0 \\ \therefore & p=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because p \in Z)\\ & \lambda=\frac{5+1}{9}=\frac{6}{9}=\frac{2}{3} . \\ \because & \frac{m}{n}=\lambda \therefore \frac{3 m+n}{3 n}=\frac{m}{n}+\frac{1}{3} \\ & =\frac{2}{3}+\frac{1}{3}=1=p \end{array} \end{aligned} $

Let $M$ be the foot of perpendicular

$ M=\left(\frac{\lambda+2}{\lambda+1}, \frac{\lambda-1}{\lambda+1}, \frac{-2 \lambda+1}{\lambda+1}\right) $

$\because \mathbf{P M} \cdot \mathbf{A B}=0$

$ \begin{aligned} & \left(\frac{\lambda+2}{\lambda+1}+1\right)(1)+\left(\frac{\lambda-1}{\lambda+1}-2\right)(-2) +\left(\frac{-2 \lambda+1}{\lambda+1}+1\right) 3=0 \\ & \begin{aligned} & \therefore 2 \lambda+3-2(-\lambda-3)+3(-\lambda+2)=0 \\ & \Rightarrow 2 \lambda+3+2 \lambda+6-3 \lambda+6=0 \Rightarrow \lambda=-15 \\ & \therefore \alpha+\beta+\gamma=\frac{\lambda+2+\lambda-1-2 \lambda+1}{\lambda+1} \\ &= \frac{2}{\lambda+1}=\frac{2}{-15+1}=\frac{-1}{7} \end{aligned} \end{aligned} $

Equation of plane containing $A(2,1,-1), B(6,-3,2) C(-3,12,4)$ is

$ \begin{aligned} & \left|\begin{array}{ccc} x-2 & y-1 & z+1 \\ 4 & -4 & 3 \\ 9 & -15 & -2 \end{array}\right|=0 \\ & \begin{array}{r} \Rightarrow(x-2)(8+45)-(y-1)(-8-27) +(z+1)(-60+36)=0 \end{array} \\ & \Rightarrow 53 x-106+35 y-35-24 z-24=0 \\ & \Rightarrow 53 x+35 y-24 z-165=0 \\ & \begin{aligned} \therefore & \frac{d}{b+c}=\frac{-165}{35-24}=\frac{-165}{11}=-15 \end{aligned} \end{aligned} $

Given, normal vector to the plane is $\mathbf{n}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$

And direction vector to the line is $\mathbf{b}=\mathbf{j}-2 \hat{\mathbf{k}}$

Since, $\mathbf{b} \cdot \mathbf{n}=(\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$

$ =3-10=-7 \neq 0 $

$\Rightarrow$ line is not perpendicular to the plane Now, $\mathbf{P}(t)=\hat{\mathbf{i}}+(-3+t) \hat{\mathbf{j}}-2 t \hat{\mathbf{k}}$

for minimum distance, $\mathbf{p}(t) \cdot \mathbf{n}=0$

$ \begin{array}{ll} \Rightarrow & 2+(-9+3 t)-10 t=0 \\ \Rightarrow & -7 t-7=0 \Rightarrow t=-1 \\ & \therefore \quad \mathbf{p}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \therefore & \quad A P=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \\ & \quad=-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{array} $

$\therefore$ Equation of plane

$ \begin{array}{rr} & (\mathbf{r}-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})) \cdot(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \\ \Rightarrow \quad & ((x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})) \\ & ((-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \\ \Rightarrow \quad & ((x-1) \hat{\mathbf{i}}+(y+4) \hat{\mathbf{j}}+(z-2 \hat{\mathbf{k}}) \\ & (-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \end{array} $

$ \begin{array}{ll} \Rightarrow & -(y+4)+2(z-2)=0 \\ \Rightarrow & -y-4+2 z-4=0 \\ \Rightarrow & -y+2 z-8=0 \\ \Rightarrow & -y+2 z=8 \\ \Rightarrow & \mathbf{r} \cdot(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=8 \end{array} $

Given, equation of planes

$ 3 x+4 y+7 z=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And $\quad x-y+z=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$\therefore$ Direction ratio of normal to the plane (i) is $3,4,7$.

and direction ratio of normal to the plane (ii) is $1,-1,1$

$\therefore$ Direction ratio of $L$,

$ \begin{aligned} \mathbf{n} & =\mathbf{n}_1 \times \mathbf{n}_2 \\ & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right| \\ & =\hat{\mathbf{i}}(4+7)-\hat{\mathbf{j}}(3-7)+\hat{\mathbf{k}}(-3-4) \\ & =11 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \end{aligned} $

$\therefore$ Direction ratio of $L$ are $11,4,-7$

Given, points $(1,1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane

$ \begin{aligned} & 3 x+4 y-12 z+13=0 \\ & \therefore \quad\left|\frac{3(1)+4(1)-12(\lambda)+13}{\sqrt{9+16+144}}\right| \\ & \quad=\left|\frac{3(-3)+4(0)-12(1)+13}{\sqrt{9+16+144}}\right| \\ & \Rightarrow \quad \frac{|20-12 \lambda|}{\sqrt{169}}=\frac{|-8|}{\sqrt{169}} \\ & \Rightarrow \quad|20-12 \lambda|=8 \end{aligned} $

Taking positive sign, we get

$ \begin{aligned} & 20-12 \lambda=8 \\ \Rightarrow \quad & -12 \lambda=-12 \Rightarrow \lambda=1 \end{aligned} $

And taking negative sign, we get

$ \begin{array}{rlrl} \Rightarrow & & 20-12 \lambda & =-8 \Rightarrow-12 \lambda=-28 \\ \Rightarrow & & 12 \lambda & =28 \Rightarrow \lambda=\frac{7}{3} \\ & \therefore & \lambda =1, \frac{7}{3} \end{array} $

Given,

$ \mathbf{r}_1=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

So, $\mathbf{a}_1=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$,

$ \mathbf{b}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

And $\mathbf{r}_2=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})+s(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$

So, $\mathbf{a}_2=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}), \mathbf{b}_2=(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$

Since, shortest distance,

$ d=\frac{\left|\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right|}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} $

Now, $\mathbf{a}_2-\mathbf{a}_1=\langle 1,2,-4\rangle-\langle 3,-5,2\rangle$

$ =\langle 1-3,2-(-5),-4-2\rangle $

$ \begin{aligned} &\begin{gathered} =<-2,7,-6> \\ \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & -1 \\ 6 & 3 & -2 \end{array}\right| \\ =(-6-(-3)) \hat{\mathbf{i}}-(-8-(-6)) \hat{\mathbf{j}}+(12-18) \hat{\mathbf{k}} \\ =-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ \left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{(-3)^2}+2^2+(-6)^2 \\ =\sqrt{9+4+36}=\sqrt{49}=7 \end{gathered}\\ &\text { So, } \quad d=\frac{|\langle-2,7,-6\rangle \cdot\langle-3,2,-6\rangle|}{7}\\ &=\frac{|6+14+36|}{7}=\frac{56}{7}=8 \end{aligned} $

$ \text { } \begin{aligned} A B & =\sqrt{(1-0)^2+(5-3)^2+(6-4)^2} \\ & =\sqrt{1+4+4}=\sqrt{9}=3 \\ A C= & \sqrt{(-2-0)^2+(0-3)^2+(-2-4)^2} \\ = & \sqrt{4+9+36}=\sqrt{49}=7 \end{aligned} $

Using the angle bisector theorem,

$ \frac{B D}{D C}=\frac{A B}{A C}=\frac{3}{7} $

Since, $D$ divides $B C$ in the ratio $3: 7$, using the section formula

$ \begin{aligned} & D=\left(\frac{7 \cdot 1+3 \cdot(-2)}{7+3}, \frac{7 \cdot 5+3 \cdot 0}{7+3}, \frac{7 \cdot 6+3 \cdot(-4)}{7+3}\right) \\ & \Rightarrow\left(\frac{7-6}{10}, \frac{35}{10}, \frac{42-6}{10}\right) \\ & \Rightarrow\left(\frac{1}{10}, \frac{35}{10}, \frac{36}{10}\right)=(0.1,35,3.6) \end{aligned} $

Now,

$ \begin{aligned} & A D=\sqrt{(0.1-0)^2+(3.5-3)^2+(3.6-4)^2} \\ & \Rightarrow \sqrt{(0.1)^2+(0.5)^2+(-0.4)^2} \\ & \Rightarrow \sqrt{0.01+0.25+0.16}=\sqrt{0.42} \\ & \Rightarrow \sqrt{\frac{42}{100}}=\frac{\sqrt{42}}{10} \end{aligned} $

$ \begin{aligned} &\text { Given two lines } 2 l+m-n=0\\ &\begin{aligned} & \Rightarrow \quad n=2 l+m \text { and } l^2-2 m^2+n^2=0 \\ & \Rightarrow \quad l^2-2 m^2+(2 l+m)^2=0 \\ & \Rightarrow \quad l^2-2 m^2+4 l^2+4 l m+m^2=0 \\ & \Rightarrow \quad 5 l^2+4 l m-m^2=0 \\ & \Rightarrow \quad 5\left(\frac{l}{m}\right)^2+4\left(\frac{l}{m}\right)-1= \\ & \text { Let } x=\frac{l}{m}, 5 x^2+4 x-1=0 \\ & \Rightarrow \quad x=\frac{-4 \pm \sqrt{4^2-4(9)(-1)}}{2(9)} \\ & \Rightarrow \quad \frac{-4 \pm \sqrt{36}}{10}=\frac{-4 \pm 6}{10} \\ & \therefore \quad x_1=\frac{2}{10} \text { and } x_2=\frac{-10}{10}=-1 \end{aligned} \end{aligned} $

Case $\mathrm{I} \frac{l}{m}=\frac{1}{5}$

$ \Rightarrow \quad m=5 l $

So, $n=2 l+m=2 l+5 l=7 l$

Direction cosine ( $l_1, m_1, n_1$ ) are proportional to $(l, 5 l, 7 l)$, i.e., $(1,5,7)$

Case II $\frac{l}{m}=-1$

$ \Rightarrow \quad m=-l $

So, $n=2 l+m=2 l-l=l$

Direction cosine ( $l_2, m_2, n_2$ ) are proportional to $(l,-l, l)$ i.e., $(1,-1,1)$

Now,

$ \begin{aligned} & \cos \theta=\frac{l_1 l_2+m_1 m_2+n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \cdot \sqrt{l_2^2+m_2^2+n_2^2}} \\ & \Rightarrow \frac{1 \times 1+5 \times(-1)+7 \times 1}{\sqrt{1^2+5^2+7^2} \cdot \sqrt{1^2+(-1)^2+1^2}} \\ & \Rightarrow \frac{1-5+7}{\sqrt{1+25+49} \cdot \sqrt{1+1+!}} \\ & \Rightarrow \frac{3}{\sqrt{75} \cdot \sqrt{3}}=\frac{3}{5 \cdot 3} \Rightarrow \frac{3}{15}=\frac{1}{5} \\ & \therefore \cos \theta=\frac{1}{5} \end{aligned} $

Let $P_1=(2,1,2)$ and $P_2=(1,2,1)$

So, $\mathbf{P}_1 \mathbf{P}_2=P_2-P_1$

$ \begin{array}{ll} \Rightarrow & (1,2,1)-(2,1,2) \\ \Rightarrow & (1-2,2-1,1-2 \\ \Rightarrow & (-1,1,-1) \end{array} $

The plane $2 x-y+2 z=1$ has a normal vector $\mathbf{n}_1=(2,-1,2)$.

Since, the required plane is perpendicular to $2 x-y+2 z=1$, its normal vector $\mathbf{n}=(a, b, c)$ must be perpendicular to $\mathbf{n}_1$.

So, $\mathbf{n}$ must be perpendicular to $\mathbf{P}_1 \mathbf{P}_2$.

$\therefore \mathbf{n}$ is parallel to the cross product of $\mathbf{P}_1 \mathbf{P}_2$ and $\mathbf{n}_1$.

$ \begin{aligned} \mathbf{n} & =\mathbf{P}_1 \mathbf{P}_2 \times \mathbf{n}_1=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{array}\right| \\ & =(2-1) \hat{\mathbf{i}}-(-2+2 \hat{\mathbf{j}}+(1-2 \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-\hat{\mathbf{k}}=\langle 1,0,-1\rangle \end{aligned} $

Now, equation of the point $(2,1,2)$ and the normal vector $(1,0,-1)$ is

$ \begin{aligned} & 1(x-2+0(y-1)-1(z-2)=0 \\ & \Rightarrow \quad x-2-z+2=0 \\ & \Rightarrow \quad x-z=0 \end{aligned} $

Comparing this with $a x+b y+c z+d=0$, we have

$ \begin{aligned} & a=1, b=0, c=-1, d=0 \\ \therefore & \frac{a+b}{c+d}=\frac{1+0}{-1+0}=-1 \end{aligned} $

Step 1: Find the Direction Vector of the Line

The line joins the points $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ (which is (1, 2, 0)) and $\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ (which is (0, 1, -2)).

So, the direction vector is found by subtracting these: $(0-1, 1-2, -2-0) = (-1, -1, -2)$.

Step 2: Write the Equation of the Line

The equation of the line is: $(x, y, z) = (1, 2, 0) + t(-1, -1, -2)$.

This gives: $x = 1-t$, $y = 2-t$, $z = -2t$.

Step 3: Find the Equation of the Plane

The plane goes through $2\hat{\mathbf{i}}-\hat{\mathbf{j}}$ (2, -1, 0), $2\hat{\mathbf{j}}+3\hat{\mathbf{k}}$ (0, 2, 3), and $\hat{\mathbf{k}}-2\hat{\mathbf{i}}$ (-2, 0, 1).

Step 4: Get Two Vectors in the Plane

Let $P = (2, -1, 0)$, $Q = (0, 2, 3)$, $R = (-2, 0, 1)$. Compute: $PQ = Q-P = (-2, 3, 3)$, $PR = R-P = (-4, 1, 1)$.

Step 5: Find the Normal Vector by Cross Product

The normal vector of the plane is $PQ \times PR$:

$ \begin{aligned} & = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & 3 & 3 \\ -4 & 1 & 1 \end{array}\right| \\ & = \hat{\mathbf{i}}(3\cdot1 - 3\cdot1) - \hat{\mathbf{j}}(-2\cdot1 - 3\cdot-4) + \hat{\mathbf{k}}(-2\cdot1 - 3\cdot-4) \\ & = \hat{\mathbf{i}}(0) - \hat{\mathbf{j}}(10) + \hat{\mathbf{k}}(10) \\ & = -10 \hat{\mathbf{j}} + 10 \hat{\mathbf{k}} \end{aligned} $

Step 6: Write Equation of the Plane

The normal vector is $(0, -10, 10)$. The plane passes through $P = (2, -1, 0)$, so the equation is: $0(x-2) - 10(y+1) + 10(z-0) = 0$.

This simplifies to $-10(y+1) + 10z = 0$, or $z - y = 1$.

Step 7: Find Where the Line Meets the Plane

Plug the line equations $y=2-t$, $z=-2t$ into the plane equation $z-y=1$:

$-2t - (2-t) = 1$

$-2t - 2 + t = 1$

$-t - 2 = 1$

$-t = 3$

$t = -3$

Step 8: Find the Intersection Point

Put $t=-3$ into the line equations:

$x = 1 - (-3) = 4$

$y = 2 - (-3) = 5$

$z = -2 \times (-3) = 6$

So, the point $\mathbf{r} = (4, 5, 6)$.

Step 9: Find $\mathbf{r} \cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$

This is $4 \times 1 + 5 \times 1 + 6 \times 1 = 15$.

A plane passing through the line of intersection of two planes

$ \begin{aligned} & \therefore \mathbf{r} \cdot((\hat{\mathbf{i}}-2 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{j}}+\hat{\mathbf{k}}))=3+5 \lambda \\ & \Rightarrow \mathbf{r} \cdot(\hat{\mathbf{i}}+2 \lambda \hat{\mathbf{j}}+(-2+\lambda) \hat{\mathbf{k}})=3+5 \lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { put } \mathbf{r}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \text { then, } 1+4 \lambda+3(-2+\lambda)=3+5 \lambda \\ & \Rightarrow \quad 1-6+7 \lambda=3+5 \lambda \Rightarrow \lambda=4 \end{aligned}\\ &\text { Substitute } \lambda \text { in Eq. (i) }\\ &\mathbf{r} \cdot(\hat{\mathbf{i}}+8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=23 \end{aligned} $

$a=(-1,2,3), B=(2,-3,1)$,

$ \begin{aligned} & C=(3,1,-2) \\ & A B=\sqrt{(2+1)^2+(-3-2)^2+(1-3)^2}=\sqrt{37} \\ & B C=\sqrt{(3-2)^2+(1+3)^2+(-2-1)^2}=\sqrt{26} \end{aligned} $

$ A C=\sqrt{(3+1)^2+(1-2)^2+(-2-3)^2}=\sqrt{42} $

$\because A B \neq B C \neq A C$ and it does not follow right angled triangle property.

Hence, $A, B, C$ form a scalene triangle.

$\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos \frac{\pi}{3}=\frac{1}{2}$

and we know, $l^2+m^2+n^2=1$

$ \begin{aligned} & \Rightarrow\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+\cos ^2 \theta=1 \\ & \Rightarrow \frac{1}{2}+\frac{1}{4}+\cos ^2 \theta=1 \Rightarrow \cos ^2 \theta=\frac{1}{4} \\ & \Rightarrow \cos \theta=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{aligned} $

Hence, direction cosines are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$

Direction ratio of normal to the plane $2 x-3 y+5 z=7$ is $2,-3,5$ and to $5 x+2 y-3 z=11$ is $\langle 5,2,-3\rangle \because$ Required plane is perpendicular to both, its normal vector $\langle 1, b, c\rangle$ must be perpendicular to both given normal vectors.

$ \therefore\langle 1, b, c\rangle \cdot\langle 2,-3,5\rangle=2-3 b+5 c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\langle 1, b, c\rangle \cdot\langle 5,2,-3\rangle$

$ =5+2 b-3 c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

By Eq. (i) and (ii), we get

$ b=-31, c=-19 $

The equation of plane is $x-31 y-19 z+d=0$ is passes through $(3,2,5)$

$ \therefore 3-62-95+d=0 \Rightarrow d=154 $

Hence, $2 b+3 c+d=2(-31) +3(-19)+154=35$

Let $A(2,-1,1), B(1,-3,-5)$ and $C(3,-4,-4)$ be vertices of $\triangle A B C$

$ \begin{aligned} \therefore \quad a & =\sqrt{4+1+1}=\sqrt{6} \\ b & =\sqrt{1+9+25}=\sqrt{35} \\ c & =\sqrt{1+4+36}=\sqrt{41} \\ \because a^2+b^2 & =c^2 \end{aligned} $

$\therefore \triangle A B C$ is right-angled triangle

∴ Circumradius $=\frac{c}{2}=\frac{\sqrt{41}}{2}$

∵ Median $A D$ is equally inclined with coordinate axes

$ A D=\frac{\lambda-5}{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}+\left(\frac{\mu-8}{2}\right) \hat{\mathbf{k}} $

Let $\alpha, \beta, \gamma$ be direction ratio of $A D \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$

$ \begin{array}{rlrl} & \because & \alpha & =\beta=\gamma \\ & & 3 \cos ^2 \alpha & =1 \Rightarrow \cos ^2 \alpha=\frac{1}{3} \\ & \therefore & \frac{\lambda-5}{2} & =1, \frac{\mu-8}{2}=1 \\ & & \lambda=7, \mu=10 \\ & \therefore & 10 \lambda-7 \mu & =0 \end{array} $

∵ Plane is $\perp^r$ to planes $x+2 y-z=1$ and $3 x-4 y+z=5$

$ \begin{aligned} \therefore \mathbf{n} & =\mathbf{n}_1 \times \mathbf{n}_2 \\ & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ 3 & -4 & 1 \end{array}\right|=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-10 \hat{\mathbf{k}} \end{aligned} $

∴ Equation of plane passing through origin and $\perp^r$ to $-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$

$ \begin{aligned} \mathbf{r} \cdot \mathbf{n} & =\mathbf{a} \cdot \mathbf{n} \\ \Rightarrow \quad-2 x-4 y-10 z & =0 \\ \Rightarrow \quad x+2 y+5 z & =0 \end{aligned} $

$\because L_1$ passes through $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ i.e., $(1,1,0)$ and $k-i$ i.e., $(-1,0,1)$

∴ Equation of $L_1$

$ \begin{aligned} & \frac{x-1}{-1-1}=\frac{y-1}{0-1}=\frac{z-0}{1-0} \\ \Rightarrow & \frac{x-1}{-2}=\frac{y-1}{-1}=\frac{z}{1} \\ \Rightarrow & \frac{x-1}{2}=\frac{y-1}{1}=\frac{z}{-1} \end{aligned} $

any point ( $2 \lambda+1, \lambda+1,-\lambda$ )

Similarly $L_2$ passes through $(0,1,2)$ and parallel $\mathbf{i}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$

⇒ Equation of $L_2$

$ \frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1} $

Any point on $L_2 \Rightarrow(\mu, \mu+1, \mu+2)$

∴ Point of intersection

$ \begin{aligned} & 2 \lambda+1=\mu, \lambda+1=\mu+1,-\lambda=\mu+2 \\ & \therefore \quad \lambda=\mu \Rightarrow \lambda=\mu=-1 \end{aligned} $

∴ Point of intersection $=(-1,0,1)$

$ \therefore y-x=1=z $

Let $P(x, y, z)$ be a point that is the same distance from each of the points $A(2,0,3)$, $B(0,3,2)$, and $C(0,0,1)$.

To find $P$, let's set all the distances equal:

Step 1: Equal Distance from $A$ and $B$

The distance from $P$ to $A$ is $\sqrt{(x-2)^2 + (y-0)^2 + (z-3)^2}$.

The distance from $P$ to $B$ is $\sqrt{(x-0)^2 + (y-3)^2 + (z-2)^2}$.

Set these distances equal (we can remove the square roots since both must be positive):

$(x-2)^2 + y^2 + (z-3)^2 = x^2 + (y-3)^2 + (z-2)^2$

Simplify Equation 1:

Expand both sides:

Left: $x^2 - 4x + 4 + y^2 + z^2 - 6z + 9$
Right: $x^2 + y^2 - 6y + 9 + z^2 - 4z + 4$

Subtract the right side from the left side:

$-4x - 6z + 4 + 9 - (-6y + 9 - 4z + 4) = 0$

$-4x - 6z + 13 + 6y - 9 + 4z - 4 = 0$

Group like terms:

$-4x + 6y - 2z = 0$

Or, $2x - 3y + z = 0$ (divide by -2)

This is Equation (i).

Step 2: Equal Distance from $B$ and $C$

Write their distances equal:

$(x-0)^2 + (y-3)^2 + (z-2)^2 = x^2 + y^2 + (z-1)^2$

Expand both sides:

Left: $x^2 + y^2 - 6y + 9 + z^2 - 4z + 4$
Right: $x^2 + y^2 + z^2 - 2z + 1$

Subtract the right side from the left side:

$-6y + 9 - 4z + 4 - (-2z + 1) = 0$

$-6y + 13 - 4z + 2z - 1 = 0$

$-6y - 2z + 12 = 0$

Or, $3y + z = 6$ (divide by -2)

This is Equation (ii).

Step 3: Solve Both Equations Together

We now solve these:

Equation (i): $2x - 3y + z = 0$
Equation (ii): $3y + z = 6$

From Equation (ii): $z = 6 - 3y$

Substitute $z$ in Equation (i):
$2x - 3y + (6 - 3y) = 0$
$2x - 6y + 6 = 0$

$2x = 6y - 6$
$x = 3y - 3$

Let's set $y = 2$ (for a possible solution):
$\to x = 3(2) - 3 = 6 - 3 = 3$
$z = 6 - 3(2) = 6 - 6 = 0$

So, the point $P$ with coordinates $(3, 2, 0)$ fits both equations.

We are given the direction ratios for $L_1$ as $1, -2, 2$.

We are also given the direction ratios for $L_2$ as $-2, 3, -6$.

To find the direction ratios for a line that is perpendicular to both $L_1$ and $L_2$, we need to take the cross product of the direction ratios of $L_1$ and $L_2$.

Let us use the formula for the cross product. If the direction ratios of the first line are $(a_1, b_1, c_1)$ and of the second line are $(a_2, b_2, c_2)$, then the perpendicular line will have direction ratios:

$(b_1c_2 - c_1b_2,\; c_1a_2 - a_1c_2,\; a_1b_2 - b_1a_2)$

Now substitute the given values:

$( -2 \times -6 - 2 \times 3,\; 2 \times -2 - 1 \times -6,\; 1 \times 3 - ( -2 ) \times -2 )$

This becomes:

$(12 - 6,\; -4 + 6,\; 3 - 4 )$

Which simplifies to:

$6, 2, -1$

So, the direction ratios for the required line are $6, 2, -1$.

Step 1: Formula for the image of a point with respect to a plane

If you have a point $A(x_1, y_1, z_1)$ and a plane $ax + by + cz + d = 0$, the image point $B(\alpha, \beta, \gamma)$ is found using the following:

$ \frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2} $

Step 2: Plug in the values

Here, $A(1,1,1)$ and the plane is $4x+2y+4z+1=0$. So, $a=4$, $b=2$, $c=4$, and $d=1$.

Calculate $ax_1 + by_1 + cz_1 + d = 4 \times 1 + 2 \times 1 + 4 \times 1 + 1 = 4 + 2 + 4 + 1 = 11$.

Calculate $a^2 + b^2 + c^2 = 4^2 + 2^2 + 4^2 = 16 + 4 + 16 = 36$.

Now, substitute both results into the formula:

$ \frac{\alpha - 1}{4} = \frac{\beta - 1}{2} = \frac{\gamma - 1}{4} = \frac{-2 \times 11}{36} $

This simplifies to:

$ \frac{\alpha - 1}{4} = \frac{\beta - 1}{2} = \frac{\gamma - 1}{4} = \frac{-22}{36} = \frac{-11}{18} $

Step 3: Solve for the coordinates

For $\alpha:$ $ \frac{\alpha - 1}{4} = \frac{-11}{18} \Rightarrow \alpha-1 = 4 \cdot \frac{-11}{18} = \frac{-44}{18} = \frac{-22}{9} \Rightarrow \alpha = 1 + \frac{-22}{9} = \frac{-13}{9} $

For $\beta:$ $ \frac{\beta - 1}{2} = \frac{-11}{18} \Rightarrow \beta-1 = 2 \cdot \frac{-11}{18} = \frac{-22}{18} = \frac{-11}{9} \Rightarrow \beta = 1 + \frac{-11}{9} = \frac{-2}{9} $

For $\gamma:$ $ \frac{\gamma - 1}{4} = \frac{-11}{18} \Rightarrow \gamma-1 = 4 \cdot \frac{-11}{18} = \frac{-44}{18} = \frac{-22}{9} \Rightarrow \gamma = 1 + \frac{-22}{9} = \frac{-13}{9} $

Step 4: Add the coordinates

$ \alpha + \beta + \gamma = \frac{-13}{9} + \frac{-2}{9} + \frac{-13}{9} = \frac{-13-2-13}{9} = \frac{-28}{9} $

Assertion Given lines $\mathbf{r}=\mathbf{a}+t \mathbf{b}$ and $\mathbf{r}=\mathbf{p}+s \mathbf{q}$

We know that the lines are coplanar if $(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})=0$

Thus, the assertion is wrong.

Reason The shortest distance between two skew lines is

$ \begin{aligned} & \qquad d=\frac{|(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})|}{|\mathbf{b} \times \mathbf{q}|} \\ & \Rightarrow \quad d|\mathbf{b} \times \mathbf{q}|=|(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})| \\ & \text { Thus, the reason is } \end{aligned} $

So, $A$ is false, but $R$ is true.

Let the two fixed points be $A(-3,1,2)$ and $B(1,-2,4)$.

Let $P(x, y, z)$ be the point at which the line segment $A B$ subtends a right angle.

Then, $\angle A P B=90^{\circ}$

$ \Rightarrow \quad \mathbf{P A} \cdot \mathbf{P B}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, $\mathbf{P A}=(-3-x, 1-y, 2-z)$

$ \mathbf{P B}=(1-x,-2-y, 4-z) $

So, $\mathbf{P A} \cdot \mathbf{P B}=(-3-x)(1-x) +(1-y)(-2-y)+(2-z)(4-z)=0 $

$ \begin{array}{cc} \Rightarrow & -3+3 x-x+x^2-2-y+2 y +y^2+8-2 z-4 z+z^2=0 \\ \Rightarrow & x^2+2 x+y^2+y+z^2-6 z+3=0 \\ \Rightarrow & x^2+y^2+z^2+2 x+y-6 z+3=0 \end{array} $

Given, vertices of $\triangle A B C$ are $A(1,2,3), B(2,3,-1)$ and $C(3,-1,-2)$

Now, $A B=B-A=(2,3,-1) -(1,2,3)=(1,1,-4) $

Now,

$ \mathbf{C B}=(3,-1,-2)-(2,3,-1)=(1,-4,-1) $

$ \begin{aligned}So,\,\, |\mathbf{A B}| & =\sqrt{1^2+1^2+(-4)^2} \\ & =\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2} \\ |\mathbf{C B}| & =\sqrt{1^2+(-4)^2+(-1)^2} \\ & =\sqrt{18}=3 \sqrt{2} \end{aligned} $

So, d.c. of $\mathbf{A B}=\frac{\mathbf{A B}}{|\mathbf{A B}|}=\frac{(1,1,-4)}{3 \sqrt{2}}$

$ =\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}\right) $

And d.c. of $\mathbf{C B}=\frac{\mathbf{C B}}{|\mathbf{C B}|}=\frac{(1,-4,-1)}{3 \sqrt{2}}$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(\frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}, \frac{-1}{3 \sqrt{2}}\right) $

$\therefore$ d.c.'s of the internal bisector of $\angle A B C$ are

$ \begin{aligned} & \left(\frac{1}{3 \sqrt{2}}+\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}-\frac{4}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}-\frac{1}{3 \sqrt{2}}\right) \\ & =\left(\frac{2}{3 \sqrt{2}}, \frac{-3}{3 \sqrt{2}}, \frac{-5}{3 \sqrt{2}}\right) \end{aligned} $

So, the D.c.'s are $\langle 2,-3,-5\rangle$.

Given points are $A(2,0,-1)$, $B(1,-2,0), C(1,2,-1)$ and $D(0,-1,-2)$.

$ \begin{aligned} & \text { So, } \begin{aligned} & \mathbf{A B}=B-A=(1,-2,0)-(2,0,-1) \\ &=(-1,-2,+1) \\ & \begin{aligned} \mathbf{A C} & =C-A=(1,2,-1)-(2,0,-1) \\ & =(-1,2,0) \end{aligned} \end{aligned} \end{aligned} $

Now, normal vector

$ \begin{aligned} & \mathbf{n}_1=\mathbf{A B} \times \mathbf{A C}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & -2 & 1 \\ -1 & 2 & 0 \end{array}\right| \\ & =(0-2) \hat{\mathbf{i}}-(0+1) \hat{\mathbf{j}}+(-2-2) \hat{\mathbf{k}} \\ & =-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-4 \hat{\mathbf{k}}=(-2,-1,-4) \\ & \text { Now, } \mathbf{A D}=D-A=(0,-1,-2)-(2,0,-1) \\ & =(-2,-1,-1) \end{aligned} $

So, normal vector $n_2=\mathbf{A C} \times \mathbf{A D}$

$ =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 2 & 0 \\ -2 & -1 & -1 \end{array}\right| $

$ \begin{aligned} & =(-2-0) \hat{\mathbf{i}}-(1+0) \hat{\mathbf{j}}+(1+4) \hat{\mathbf{k}} \\ & =-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}=(-2,-1,5) \end{aligned} $

So, $\mathbf{n}_1 \times \mathbf{n}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & -1 & -4 \\ -2 & -1 & 5\end{array}\right|$

$ \begin{gathered} =(-5-4) \hat{\mathbf{i}}-(-10-8) \hat{\mathbf{j}}+(2-2) \hat{\mathbf{k}} \\ =-9 \hat{\mathbf{i}}+18 \hat{\mathbf{j}}+0 \hat{\mathbf{k}} \\ (-9,18,10) \end{gathered} $

And $\mathbf{n}_1 \cdot \mathbf{n}_2=(-2,-1,-4) \cdot(-2,-1,5)$

$ =4+1-20=-15 $

So, the tangent of the angle between the two planes

$ \begin{aligned} \tan \theta & =\frac{\left|\mathbf{n}_1 \times \mathbf{n}_2\right|}{\left|\mathbf{n}_1 \cdot \mathbf{n}_2\right|}=\frac{\sqrt{(-9)^2+(18)^2+0^2}}{|-15|} \\ & =\frac{\sqrt{81+324+0}}{|-15|}=\frac{\sqrt{405}}{15} \\ & =\frac{9 \sqrt{5}}{15}=\frac{3 \sqrt{5}}{5}=\frac{3}{\sqrt{5}} \end{aligned} $

So, the tangent of the acute angle between the two planes is $\frac{3}{\sqrt{5}}$.

$A(0,1,2), B(2,-1,3), C(1,-3,1)$

$ \begin{array}{ll} \because & \mathbf{B A}=<-2,2,-1 \geqslant \\ & \mathbf{B C}=<-1,-2,-2> \\ \because & \mathbf{B A} \cdot \mathbf{B C}=(-2)(-1)+2(-2) \\ & +(-1)(-2)=0 \end{array} $

So, $\triangle A B C$ is right angled at $B$.

$\therefore$ Orthocenter $(H)$ is at point $B=(2,-1,3)$ and circumcenter

$ (O)=\left(\frac{0+1}{2}, \frac{-3+1}{2}, \frac{1+2}{2}\right)=\left(\frac{1}{2},-1, \frac{3}{2}\right) $

Hence,

$ \begin{aligned} O H & =\sqrt{\left(2-\frac{1}{2}\right)^2+(-1+1)^2+\left(3-\frac{3}{2}\right)^2} \\ & =\sqrt{\frac{9}{4}+\frac{9}{4}}=\frac{3}{\sqrt{2}} \end{aligned} $

$\because l-2 m+n=0 \Rightarrow l=2 m-n$

Substitute in $l m+10 m n-2 n l=0$

$ \begin{aligned} & \Rightarrow(2 m-n) m+100 m n-2 n(2 m-n)=0 \\ & \Rightarrow 2 m^2+5 m n+2 n^2=0 \\ & \Rightarrow m=\frac{-5 n \pm 3 n}{4}=\frac{-n}{2},-2 n \end{aligned} $

If $m=\frac{-n}{2}, l=-2 n$

Then, direction ratios of are

$ 1: m: n=-2 n: \frac{n}{2}: n=-4:-1: 2 $

and direction cosine are

$ \begin{array}{r} (-4 k)^2+(-k)^2+(2 k)^2=1 \Rightarrow k=\frac{1}{\sqrt{21}} \\ =\left(\frac{-4}{\sqrt{21}}, \frac{-1}{\sqrt{21}}, \frac{2}{\sqrt{21}}\right) \end{array} $

If $m=-2 n$, then $l=-5 n$ and DR's are $l: m: n=-5:-2: 1$ and DC's are $(-5 k)^2+(-2 k)^2+k^2=1$

$ \Rightarrow \quad k=\frac{1}{\sqrt{30}}=\left(\frac{-5}{\sqrt{30}}, \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}\right) $

Therefore angle between given lines are

$ \begin{aligned} & \cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right| \\ & =\left\lvert\, \begin{array}{c} \frac{-4}{\sqrt{21}} \times\left(\frac{-5}{\sqrt{30}}\right)+\left(\frac{-1}{\sqrt{21}}\right) \\ \left(\frac{-2}{\sqrt{30}}\right)+\left(\frac{2}{\sqrt{21}}\right)\left(\frac{1}{\sqrt{30}}\right) \end{array}\right. \end{aligned} $

$ \begin{aligned} & =\frac{24}{\sqrt{630}}=\frac{24}{3 \sqrt{70}} \\ & =\frac{8}{\sqrt{70}} \end{aligned} $

As given, the vector from the origin to the point $(2,-1,3)$ is normal to plane.

$ \therefore \quad \mathbf{n}=<2,-1,3> $

Then, general equation of plane

$ 2 x-y+3 z=d $

$\because \quad(2,-1,3)$ lies on the plane.

$ \therefore \quad 2(2)-(-1)+3 \times 3=d \Rightarrow d=14 $

Hence, equation of the plane

$ 2 x-y+3 z-14=0 $

Given, $A=(2,-1,1), B=(2,5,1)$, $C=(0,-2,3)$

$ \begin{aligned} &\text { Using angle bisector theorem, }\\ &\frac{A B}{A C}=\frac{B D}{D C} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{\sqrt{(2-2)^2+(5-(-1))^2+(1-1)^2}}{\sqrt{(0-2)^2+(-2+1)^2+(3-1)^2}}=\frac{B D}{D C} \\ & \Rightarrow \frac{B D}{D C}=\frac{6}{3}=\frac{2}{1} \\ & \therefore B D: D C=2: 1 \end{aligned} $

Now, we find coordinates of $D$ by the help of section formula,

$ \begin{aligned} & D=\left(\begin{array}{rr} \frac{2 \times 0+1 \times 2}{2+1}, & \frac{2 \times(-2)+1 \times 5}{2+1}, \\ & \frac{2 \times 3+1 \times 1}{2+1} \end{array}\right) \\ & =\left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}\right) \end{aligned} $

Hence,

$ \begin{aligned} A D & =\sqrt{\left(2-\frac{2}{3}\right)^2+\left(-1-\frac{1}{3}\right)^2+\left(1-\frac{7}{3}\right)^2} \\ & =\sqrt{\left(\frac{4}{3}\right)^2+\left(\frac{-4}{3}\right)^2+\left(\frac{-4}{3}\right)^2} \\ & =\sqrt{\frac{48}{9}}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}} \end{aligned} $

Given equation of plane $\pi$ is

$ a x+b y+11 z+d=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, the normal vector of $\pi$ is

$ \mathbf{n}_1=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+11 \hat{\mathbf{k}}=\langle a, b, 11\rangle $

Normal of first plane:

$ \mathbf{n}_2=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+1 \cdot \hat{\mathbf{k}}=\langle 2,-3,1\rangle $

Normal of second plane:

$ \mathbf{n}_3=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\langle 3,1,-1\rangle $

According to question, if plane $\pi$ is perpendicular to both then its normal vector is perpendicular to the normals of the two planes.

So, $\mathbf{n}_1 \cdot \mathbf{n}_2=0 \Rightarrow 2 a-3 b+11=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \mathbf{n}_1 \cdot \mathbf{n}_3=0 \Rightarrow 3 a+b-11=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On solving Eqs. (ii) and (iii) we get,

$ \begin{aligned} & a=2 \text { and } b=11-3 a=11-6=5 \\ \therefore & < a, b, 11 >=< 2,5,11 > \end{aligned} $

$\therefore$ Equation of plane will become,

$ 2 x+5 y+11 z+d=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Now, distance from origin $(0,0,0)$ to the plane (iv) is

$ \begin{aligned} & \frac{|d|}{\sqrt{2^2+5^2+11^2}}=\sqrt{6} \\ \Rightarrow \quad & |d|=\sqrt{6}+\sqrt{150}=30 \\ \Rightarrow \quad & d= \pm 30 \end{aligned} $

(given)

As question demand is all intercepts (i.e. $x, y \times z$ intercepts) are positive.

So, $d$ must be negative because $x$-intercept $=\frac{-d}{2}>0 \Rightarrow d<0 y$-intercept $=-\frac{d}{5}>0 \Rightarrow d<0$

$z$-intercept $=\frac{-d}{11}>0 \Rightarrow d<0$

Hence, $d=-30=-3 \times 2 \times 5=-3 a b$

$ \begin{aligned} & \text {Point }(-\hat{\mathbf{i}}+p \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \Rightarrow \text { Point } \\ & \text { (-1, p, -3) } \\ & \text { Plane r } \cdot(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})=7 \\ & \Rightarrow 2 x-3 y+6 z-7=0 \end{aligned} $

$ \begin{aligned} P Q & =\frac{|-2-3 p-18-7|}{\sqrt{2^2+(-3)^2+6^2}} \\ 6 & =\frac{|-3 p-27|}{\sqrt{49}} \\ 42 & =|-3 p-27| \\ \Rightarrow & (-3 p-27)= \pm 42 \\ \Rightarrow & -3 p-27=42 \\ \Rightarrow & -3 p=69 \\ \Rightarrow \quad & p=-23 \\ \Rightarrow & -3 p-27=-42 \\ \Rightarrow & -3 p=-15 \\ & p=5 \end{aligned} $

Let $P$ divide $A B$ in the ratio $m: n$.

$ \begin{aligned} & 0=\frac{2 n-m}{m+n},-7=\frac{-5 n-8 m}{m+n}, 1=\frac{3 n}{m+n} \\ & \Rightarrow 2 n-m=0 \Rightarrow m=2 n \end{aligned} $

$\Rightarrow P$ divides $A B$ in the ratio $2: 1$

Since $Q$ is the hormonic conjugate of $P$ it divides $A B$ in the ratio $-2: 1$

Using the section formula with the ratio -2:1

$ \begin{aligned} & \alpha=\frac{2-2 \times(-1)}{1-2}=-4 \\ & \beta=\frac{-5-2(-8)}{1-2}=-11 \\ & \gamma=\frac{3-2(0)}{1-2}=\frac{3}{-1}=-3 \end{aligned} $

Thus, $Q=(-4,-11,-3)$

$ \Rightarrow \quad \alpha-\beta+\gamma=-4+11-3=4 $

Let $B$ divide $A C$ in the ratio $\lambda{: 1}$ The coordinates of $B$ are given by

$ \begin{aligned} x_B & =\frac{\lambda\left(x_1+k l\right)+1 x_1}{\lambda+1} \\ y_B & =\frac{\lambda\left(y_1+k m\right)+1 y_1}{\lambda+1} \\ Z_B & =\frac{\lambda\left(z_1+k n\right)+1 z_1}{\lambda+1} \\ \Rightarrow x_1+4 k l & =\frac{\lambda\left(x_1+k l\right)+x_1}{\lambda+1} \\ \Rightarrow y_1+4 k m & =\frac{\lambda\left(y_1+k m\right)+y_1}{\lambda+1} \\ \Rightarrow z_1+4 k n & =\frac{\lambda\left(z_1+k n\right)+z_1}{\lambda+1} \end{aligned} $

$ \begin{aligned} & \text { Now, }\left(x_1+4 k l\right)(\lambda+1)=\lambda\left(x_1+k l\right)+x_1 \\ & \\ \Rightarrow& x_1 \lambda+x_1+4 k l \lambda+4 k l=x_1 \lambda+k l \lambda+x_1 \\ & \Rightarrow \quad 4 k l \lambda+4 k l=k l \lambda \\ & \Rightarrow \quad \lambda=\frac{4 k l}{-3 k l}=\frac{4}{-3} \Rightarrow \text { Ratio }=4:-3 \end{aligned} $

$(2,3,1)(1,3,2)$

$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(6-3)-\hat{\mathbf{j}}(4-1)+\hat{\mathbf{k}}(6-3) \\ & =3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

Since, lines of intersection of the given planes makes an angle $\alpha$ with positive $X$-axis.

$ \therefore \cos \alpha=\frac{3}{\sqrt{3^2+(-3)^2+3^2}}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} $

Equation of line,

$ r=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Plane passes through $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and is parallel to the vectors $\mathbf{v}_1=2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$,

$ \mathbf{v}_2=\hat{\mathbf{i}}+2 \hat{\mathbf{k}} $

Any point on the plane can be written as

$ \mathbf{r}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+x(2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+v(\hat{\mathbf{i}}+2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii),

$ \begin{aligned} & 1+2 \lambda=1+V \Rightarrow 2 \lambda=V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & x=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \lambda=-x+2 V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{aligned} $

We get $\lambda=-\frac{2}{3}$

$ \begin{aligned} \mathbf{r} & =(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})-\frac{2}{3}(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}) \\ & =\frac{-1}{3}(\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} & |\mathbf{O D}|=\sqrt{0^2+0^2+a^2}=a \\ & |\mathbf{O C}|=\sqrt{a^2+a^2+a^2}=a \sqrt{3} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \cos \alpha & =\frac{a^2}{a \sqrt{3} \cdot a}=\frac{1}{\sqrt{3}} \\ \Rightarrow \quad \alpha & =\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ |O E| & =a \sqrt{2} \\ \cos \theta & =\frac{2 a^2}{\sqrt{2} \cdot \sqrt{3} a^2} \Rightarrow \theta=\cos ^{-1} \sqrt{\frac{2}{3}} \end{aligned} $

Plane passing through $A(1,-2,3)$ and $B(6,4,5)$

Normal vector of the given plane $\pi$

$ \mathbf{n}_1=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Normal vector of plane

$ \begin{aligned} \mathbf{n} & =\mathbf{A B} \times \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & 6 & 2 \\ 3 & -1 & 1 \end{array}\right| \\ & =8 \hat{\mathbf{i}}+\hat{\mathbf{j}}-23 \hat{\mathbf{k}} \end{aligned} $

Equation of plane $\pi$

$ \begin{aligned} & 8(x-1)+1(y+2)-23(z-3)=0 \\ & \Rightarrow 8 x+y-23 z+63=0 \end{aligned} $

Perpendicular distances from $(0,0,0)$ to the plane $\pi$

$ d=\frac{63}{\sqrt{64+1+529}}=\frac{63}{\sqrt{594}} $

Let $P$ and $Q$ be the points on the given lines where they intersect.

$ \begin{aligned} & P=(1+t,-6+2 t, 2+t) \text { and } \\ & Q=(2 s, 4+s, 1+2 s) \end{aligned} $

Now, $P$ and $Q$ are identical

$ \begin{aligned} \therefore \quad t+1 & =2 s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -6+2 t & =4+s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 2+t & =1+2 s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On solving Eqs. (i) and (ii), we get

$ s=4 \text { and } t=7 $

Which satisfy the Eq. (iii)

$\therefore$ Point of intersection $(8,8,9)$

$ =8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}} $

Let four points $A(6,2,4), B(1,3,5)$, $C(1,-2,3)$ and $D(6, k, 2)$

$ \begin{aligned} & \mathbf{A B}=-5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{A C}=-5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{A D}=(k-2) \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Given, four points $A B C$ and $D$ are coplanar

$ \begin{aligned} & \therefore[\mathbf{A B}, \mathbf{A C}, \mathbf{A D}]=0 \\ & \Rightarrow\left|\begin{array}{ccc} -5 & 1 & 1 \\ -5 & -4 & -1 \\ 0 & k-2 & -2 \end{array}\right|=0 \\ & \Rightarrow-5(8+k-2)-1(10)+1(-5 k+10)=0 \\ & \Rightarrow-5(6+k)-10-5 k+10=0 \\ & \Rightarrow-30-5 k-10-5 k+10=0 \\ & \Rightarrow-10 k-30=0 \\ & \therefore k=-3 \end{aligned} $